Derivatives > How to find critical numbers

A **critical number** is a number “c” that either:

- Makes the derivative equal to zero, or
- Results in an undefined derivative (i.e. it’s not differentiable at that point).

Critical numbers indicate where a **change **is taking place on a graph. For example:

- An increasing to decreasing point (e.g. a local maximum), or
- A decreasing to increasing point (e.g. a local minimum).

The number “c” also has to be in the domain of the original function (the one you took the derivative of).

## How to find critical numbers

**Finding critical numbers** is relatively easy if your algebra skills are strong; Unfortunately, **if you have weak algebra skills you might have trouble finding critical numbers**. Why? Because each function is different, and algebra skills will help you to spot *undefined domain* possibilities like division by zero. If your algebra isn’t up to par—now is the time to restudy the old rules.

**Example question:** Find the critical numbers for the following function:

Step 1: **Take the derivative of the function**. Which rule you use depends upon your function type. For this example, you have a division, so use the quotient rule to get:

Step 2: **Figure out where the derivative equals zero.** This is where a little algebra knowledge comes in handy, as each function is going to be different.

For this particular function, the derivative equals zero when -18x = 0 (making the numerator zero), so one critical number for x is 0 (because -18(0) = 0).

Another set of critical numbers can be found by setting the denominator equal to zero; When you do that, you’ll find out where the derivative is undefined:

- (x
^{2}– 9) = 0 - (x – 3)(x + 3) = 0
- x = ±3

Step 3: **Plug any critical numbers you found in Step 2 into your original function** to check that they are in the domain of the original function. For this function, the critical numbers were 0, -3 and 3. Let’s plug in 0 first and see what happens:

f(x) = ^{02}⁄_{02-9} = 0. Therefore, 0 is a critical number.

For +3 or -3, if you try to put these into the denominator of the original function, you’ll get **division by zero**, which is **undefined**.

f(x) = ^{32}⁄_{32-9} = 9/0. Therefore, 3 is not a critical number.

That means these numbers are not in the domain of the original function and are not critical numbers.

*That’s it!*

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**Stephanie Glen**. "How to Find Critical Numbers (Points)" From

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